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Scaling of the Dirac Delta (Impulse Function)

$ \displaystyle\delta(\alpha f)=\frac{1}{\alpha}\delta(f) $ $ \textstyle{ for }\alpha>0 $

Mini Proof

$ \int_{-\infty}^{\infty}\delta(x)dx = 1 $

Let $ \displaystyle y=\alpha x $

$ dx=\frac{dy}{\alpha} $

$ \displaystyle\int_{-\infty}^{\infty}\delta(\alpha x)dx=\int_{-\infty}^{\infty}\delta(y)\frac{dy}{\alpha}=\frac{1}{\alpha} $

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Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva